`100mL CuSO_(4)` (aq) was electrolyzed using inert electrodes by passing `0.965A` till the pH of the resulting solution was 1. the soluton after electrolysis was neutralized treated with excess `KI` and titrated with `0.04M Na_(2)S_(2)O_(3)`. Volume of `Na_(2)S_(2)O_(3)` required was 35mL. Assuming no volume change during electrolysis calculate:
(a) during of electrolysis if current efficiency is `80%`
(b) initial concentration `(M)` of `CuSO_(4)`.

Answer

`1250s, 0.064M`

Explanation

`pH = 1, [H^(+)] = 10^(-1)` mole/lit
`[H^(+)]` in `100mL =10^(-2)` mole
at anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)`
at anode: `Cu^(+2) +2e^(-) rarr Cu`
To produce `0.01` mole `H^(+)`, we need `0.01` Faraday.
`0.01 = (0.965 xx 8 xxt)/(96500) rArr t = 1250 sec`.
eq. of `Cu^(+2)` consumed `=0.01`
moles of `Cu^(+2)` consumed
`=(0.01)/(2) = 0.005`
m moles of `I_(2) = (0.04 xx 35)/(2)`
m moles of `CuSO_(4) = 0.04 xx 35 = 1.4`
total moles `=0.005 +0.0014 = 0.0064`
volume `=100mL = 0.1L`
`[CuSO_(4)] = (0.0064)/(0.1) = 0.064M`